Problem: Simplify the following expression: $y = \dfrac{-4x^2+3x+10}{x - 2}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(10)} &=& -40 \\ {a} + {b} &=& &=& {3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-40$ and add them together. Remember, since $-40$ is negative, one of the factors must be negative. The factors that add up to ${3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-5})({8}) &=& -40 \\ {a} + {b} &=& {-5} + {8} &=& 3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-4}x^2 {-5}x) + ({8}x +{10}) $ Factor out the common factors: $ x(-4x - 5) - 2(-4x - 5)$ Now factor out $(-4x - 5)$ $ (-4x - 5)(x - 2)$ The original expression can therefore be written: $ \dfrac{(-4x - 5)(x - 2)}{x - 2}$ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ This leaves us with $-4x - 5; x \neq 2$.